Question: The radius of the base of a cone is increasing at a rate of $10$ meters per second. The height of the cone is fixed at $6$ meters. At a certain instant, the radius is $1$ meter. What is the rate of change of the volume of the cone at that instant (in cubic meters per second)? Choose 1 answer: Choose 1 answer: (Choice A) A $20\pi$ (Choice B) B $40\pi$ (Choice C) C $200\pi$ (Choice D) D $400\pi$ The volume of a cone with radius $r$ and height $h$ is $\pi r^2\dfrac{h}{3}$.
Setting up the math Let... $r(t)$ denote the cone's radius at time $t$, and $V(t)$ denote the cone's volume at time $t$. We are given that $r'(t)=10$ and that $r(t_0)=1$ for a specific time $t_0$. We are also given that the cone's height is $6$ meters. We want to find $V'(t_0)$. Relating the measures $V(t)$ and $r(t)$ relate to each other through the formula for the volume of a cone: $V(t)=\pi[r(t)]^2\dfrac{h}{3}$ Plugging $h=6$, we get the following equation: $V(t)=2\pi[r(t)]^2$ We can differentiate both sides to find an expression for $V'(t)$ : $V'(t)=4\pi r(t)r'(t)$ Using the information to solve Let's plug ${r(t_0)}={1}$ and ${r'(t_0)}={10}$ into the expression for $V'(t_0)$ : $\begin{aligned} V'(t_0)&=4\pi{r(t_0)}{r'(t_0)} \\\\ &=4\pi({1})({10}) \\\\ &=40\pi \end{aligned}$ In conclusion, the rate of change of the volume of the cone at that instant is $40\pi$ cubic meters per second. Since the rate of change is positive, we know that the volume is increasing.